Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void amin(T &a, const T &b) {
  if (b < a) a = b;
}
template <class T>
inline void amax(T &a, const T &b) {
  if (a < b) a = b;
}
int N, M, Q;
vector<int> G[100011], P[100011];
vector<int> add(vector<int> a, const vector<int> &b) {
  a.insert(a.end(), b.begin(), b.end());
  sort(a.begin(), a.end());
  if (a.size() > 10u) a.resize(10);
  return a;
}
int per[100011][20];
int dep[100011];
vector<int> me[100011][20];
vector<int> peo;
int lca(int v, int w) {
  if (dep[v] > dep[w]) swap(v, w);
  for (int i = 0; i < 20; i++) {
    if ((dep[w] - dep[v]) & (1 << i)) {
      peo = add(peo, me[w][i]);
      w = per[w][i];
    }
  }
  if (w == v) {
    peo = add(peo, me[w][0]);
    return w;
  }
  for (int i = 19; i >= 0; i--) {
    if (per[w][i] != per[v][i]) {
      peo = add(peo, me[w][i]);
      peo = add(peo, me[v][i]);
      w = per[w][i];
      v = per[v][i];
    }
  }
  peo = add(peo, me[v][0]);
  peo = add(peo, me[w][1]);
  return per[v][0];
}
int main() {
  scanf("%d%d%d", &N, &M, &Q);
  for (int i = 0, i_len = (N - 1); i < i_len; ++i) {
    int v, w;
    scanf("%d%d", &v, &w);
    v--;
    w--;
    G[v].push_back(w);
    G[w].push_back(v);
  }
  for (int i = 0, i_len = (M); i < i_len; ++i) {
    int c;
    scanf("%d", &c);
    c--;
    P[c].push_back(i);
  }
  for (int i = 0, i_len = (N); i < i_len; ++i) {
    sort(P[i].begin(), P[i].end());
    if (P[i].size() > 10u) P[i].resize(10);
  }
  vector<int> ord;
  ord.reserve(0);
  ord.push_back(0);
  per[0][0] = 0;
  for (int i = 0, i_len = (N); i < i_len; ++i) {
    int v = ord[i];
    me[v][0] = P[v];
    for (__typeof((G[v]).begin()) e = (G[v]).begin(), e_end = (G[v]).end();
         e != e_end; ++e)
      if (*e != per[v][0]) {
        dep[*e] = dep[v] + 1;
        per[*e][0] = v;
        ord.push_back(*e);
      }
  }
  for (int t = 0, t_len = (19); t < t_len; ++t)
    for (int v = 0, v_len = (N); v < v_len; ++v) {
      per[v][t + 1] = per[per[v][t]][t];
      me[v][t + 1] = add(me[v][t], me[per[v][t]][t]);
    }
  for (int $ = 0, $_len = (Q); $ < $_len; ++$) {
    int v, w, a;
    scanf("%d%d%d", &v, &w, &a);
    v--;
    w--;
    peo.clear();
    int z = lca(v, w);
    if ((int)peo.size() > a) peo.resize(a);
    printf("%d", (int)peo.size());
    for (int i = 0, i_len = (peo.size()); i < i_len; ++i)
      printf(" %d", peo[i] + 1);
    puts("");
  }
  return 0;
}